Question

Each α particle in a beam of α particles has a kinetic energy of 7.0 MeV. Through what potential difference would you have to accelerate these α particles in order that they would have enough energy so that if one is fired head-on at a gold nucleus it could reach a point 1.1  10^-14 m from the center of the nucleus?

Answer 1

See the diagram :

Given :

Initial Kinetic Energy of alpha particle = K1 = 7 MeV

Atomic number of gold = Z = 79

charge on alpha particle = q1 = 2e

charge on gold nucleus = q2 = Ze = 79e

Final distance between alpha particle and gold nucleus (where Kinetic energy of alpha particle = 0) = r = 1.1 x 10-14 m.

Accelerated potential difference applied = V = to be calculated.

Now,

When a charged particle moves towards another charged particle of same nature, it's potential energy increases.

Electrostatic potential energy between two point charges particles is given by = .

When the alpha particle having some kinetic energy is further accelerated with potential difference V, it's new Kinetic energy = K'.

K' = K1 + q1V = K1 + (2e x V) = (7 x 106 x 1.6 x 10-19) + (2 x 1.6 x 10-19 x V)

Now, when this alpha particle moves towards the gold nucleus , it loses it's kinetic energy and gains electrostatic potential energy and finally somes to a point when total kinetic energy of the alpha particle becomes zero and converted into potential energy. This distance is termed as distance of closest approach.

the closest distance here is r (value is given).

Electrostatic potential energy of alpha particle at this point =

So, applying conservation of mechanical energy :

=> V = 6.8 x 106 V.